Chapter 1 1.0 Introduction 1.6 Energy 1.1 Scientific Method … · 1.4 Atoms and Molecules 1.10...

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Chapter 1 The Early Experiments

1.0 Introduction 1.6 Energy 1.1 Scientific Method 1.7 Electromagnetism and Coulomb’s Law 1.2 Lavoisier and the Birth of Modern Chemistry 1.8 Atomic Structure 1.3 John Dalton and Atomic Theory 1.9 Subatomic Particles, Isotopes, and Ions 1.4 Atoms and Molecules 1.10 Dimitri Mendeleev and the Periodic Law 1.5 The Mole and Molar Mass 1.11 Chapter Summary and Objectives

1.12 Exercises

1.0 INTRODUCTION

Chemistry is the science of matter, its properties, and the changes it undergoes. Chemists seek to understand our material universe at a molecular level and to use this understanding to improve our interaction with it, often creating new products that enhance our lives. These products include pharmaceuticals, fuels, plastics, batteries, soaps, perfumes, foods, fertilizers and pesticides, to name just a few. Chemists often design these products by considering the properties of the desired substance and then proposing reactions of atoms or molecules that might yield the substances of choice. This design process involves particles and processes chemists can envision but cannot see. Chemists can observe the results of a reaction, such as a color change, the formation of a gas or a solid, or the formation of the substance with the desired properties. However, they cannot view directly the collisions of the atoms or molecules in a reaction or the changes these collisions produce. Yet, chemists are confident that these collisions and changes do occur. How did we get to the point where we can envision these invisible processes? How do we know the nature of these invisible atoms and molecules? We start our study of chemistry by beginning to answer these two questions.

THE OBJECTIVES OF CHAPTER 1 ARE TO: • define the scientific method and illustrate its importance in scientific discovery;

• introduce Coulomb’s law and the electromagnetic force;

• introduce the early scientists and the experiments that eventually led to the description of the nuclear atom and its components;

• describe the subatomic particles and their characteristics; and

• explain the use of the Periodic Table in predicting an element’s physical and chemical properties.

Chapter 1 The Early Experiments 1 Copyright © by North Carolina State University

Chapter 1 The Early Experiments 2

1.1 SCIENTIFIC METHOD

Chemistry is a science, which means that all chemical knowledge is gained by the application of a set of principles and procedures known as the scientific method. To understand how chemistry progressed to where it is today and how it will progress in the future, you must first understand how this method is applied. The scientific method involves the following steps:

* Quantitative observations involve numbers, while qualitative observations do not. For example, “the mass of the object is 3.2 g” is a quantitative observation, while “the object is black” is a qualitative observation.

1. observation and collection of quantitative or qualitative data;*

2. formulation of a hypothesis to explain the observation;

3. prediction based on the hypothesis; and

4. testing the prediction.

If a test supports a hypothesis, another experiment is devised to further test the hypothesis. If a test does not support a hypothesis, then the hypothesis is changed or even discarded depending upon how badly it fails the test. After a hypothesis has been supported by many independent observers, it becomes a law or a theory. A law summarizes many observations, while a theory provides an explanation for them. Theories cannot be proven and are valid only as long as they are supported or, at least, not disproved by experiment. Our understanding is constantly evolving as the sophistication of our tools and instruments improves and our scientific knowledge increases. Many long held theories have eventually failed the test of experiment and have been modified or discarded entirely. Thus, chemistry is a dynamic science that continues to build upon past observations and theories by exploring new discoveries and hypotheses. The struggle of the earliest scientists to understand and explain the nature of things around them is an amazing journey of discovery. In this chapter, we examine some of the experiments that brought us from the birth of modern chemistry at the end of the 18th century to the discovery of subatomic particles near the end of the 19th century.

1.2 LAVOISIER AND THE BIRTH OF MODERN CHEMISTRY

Controlling fire was a major factor in the rise of humans, so it is not surprising that one of the first theories dealt with burning objects. We can make two obvious observations about fire: it is hot; and the flame leaps from the fuel. Phlogiston theory was born from little more than this kind of fireside observation. According to phlogiston theory, materials that burned contained a substance called phlogiston, and burning was thought to be the release

Copyright © by North Carolina State University

of phlogiston (the flame leaping from the fuel). The residue that remained after burning was called calx. A phlogistonist would view the brilliant white light that is produced when magnesium burns as the escape of phlogiston and represented the burning as

Magnesium → ‘calx of Magnesium’ + Phlogiston

Thus, it was believed that the metal lost phlogiston when it burned. Early in the eighteenth century, however, Antoine Lavoisier, a French chemist, showed that the mass of the calx was greater than the mass of the metal. Scientists that supported the phlogiston theory reasoned that either mass was not relevant to chemistry or that phlogiston had negative mass, but most scientists realized that phlogiston theory had failed an important test, and it was eventually discarded.

Lavoisier measured the mass of many reactions and observed that the total mass does not change. He summarized his results in the law of conservation of mass.

Law of Conservation of Mass: during a chemical reaction, the total mass (reactants + products) remains constant; that is, mass is neither created nor destroyed during a chemical reaction.

After careful and repeated experiments, he realized that burning a metal was the combination of the metal with oxygen, not the release of phlogiston. Lavoisier introduced a new way of thinking about chemistry and is known as the father of modern chemistry. He viewed the calx of a metal as the metal oxide and the reaction as

Magnesium + Oxygen → Magnesium oxide

Lavoisier was also the first to classify matter as elements or compounds. An element is a pure substance that cannot be broken down by chemical means to a simpler substance. Magnesium and oxygen are examples of elements. Today, there are over 100 known elements. A compound is a pure substance that consists of more than one element. Magnesium oxide is a compound that is formed by the combination of two elements (magnesium and oxygen).

The scientists of the early 19th century had a new system and a new way of thinking about matter as the field of chemistry was born. They began testing the concept that matter consisted of elements and compounds and that mass was indeed relevant to chemistry. After a great number of measurements of relative masses had been performed, two more laws that summarized the results were accepted.



Law of Definite Proportions: the elements of a compound are present in definite (fixed) proportions by mass. For example, the mass of table salt (sodium chloride) is always 39% sodium and 61% chlorine and that of water is always 11% hydrogen and 89% oxygen. Law of Multiple Proportions: when two different compounds are formed from the same two elements, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers. For example, water and hydrogen peroxide are both compounds that are composed only of the elements hydrogen and oxygen. There are eight grams of oxygen for each gram of hydrogen in water, but there are 16 g of oxygen for each gram of hydrogen in hydrogen peroxide. For a specified mass of hydrogen (one gram), the mass ratio of oxygen in the two compounds is 8:16 or 1:2, a ratio of small whole numbers.

Example 1.1 Sodium (Na) and oxygen (O) form two different compounds that are 59% and 74% Na by mass. Show that these compounds obey the law of multiple proportions. First, determine the mass of sodium that is combined with a specified mass of oxygen. Percents can be converted easily into grams by assuming a total mass of 100 g. For example, 59% of a 100 g sample is 59 g. The compounds consist only of Na and O, so the sum of the percents must be 100. Consequently, %O = 100 - %Na.

Next, specify a fixed mass of one of the substances, which is usually set at 1 g. In the following, it is the mass of oxygen that is fixed. The mass of Na combined with 1 g of O is obtained by dividing the mass of Na by the mass of O with which it is combined. The following table shows the results.

Compound %Na %O grams Na/1 gram O

I 59 41 59 g Na 1.4 g Na= 41 g O 1 g O

II 74 26 74 g Na 2.8 g Na = 26 g O 1 g O

Finally, determine the ratio of the masses of Na combined with 1 g O in the two compounds. The ratio of compound II to compound I is

2.8 g Na in cmpd II g Oratio = 1.4 g Na in cmpd I

g O

* 2.8 g Na in cmpd II 2.0 g Na in cmpd II = = 1.4 g Na in cmpd I 1.0 g Na in cmpd I

*

The ratio is a ratio of small whole numbers (2:1), so these compounds obey the law of multiple proportions. The ratio implies that there is twice as much Na per gram of O in compound II as there is in compound I. In fact, compound I is Na2O2 (sodium peroxide) and compound II is Na2O (sodium oxide).

* Note that the units “g O” are in the denominators, so they cancel in the ratio to yield the desired ratio of masses of Na. Units are very important and their use in solving problems will be examined in more detail later in the chapter.


1.3 JOHN DALTON AND ATOMIC THEORY (1804)

Laws hold the key to understanding nature’s secrets, and theories are our attempt to unlock the secrets. The chemists of the early 19th century had three laws to explain: conservation of mass, definite proportions, and multiple proportions. Elements and compounds were the accepted forms of matter; but what were the mass relationships telling them? In 1804, John Dalton, an English chemist, suggested an answer: elements consisted of tiny spheres, called atoms, which he likened to billiard balls with hooks on them. He assigned the following properties to atoms to assure that they behaved in a manner consistent with the laws of conservation of mass, definite proportions, and multiple proportions:

1. An element is composed of extremely small particles called atoms. The atoms of a given element all exhibit identical chemical properties,* but atoms of different elements have different chemical properties.

2. In the course of a chemical reaction, no atom disappears or is changed into another atom. This property explains The law of conservation of mass and is the basis for writing balanced chemical equations. In a balanced chemical equation, the number of each kind of atom must be the same on both sides of the equation.

3. Compounds are formed when atoms of different elements combine. In a given pure compound, the relative numbers of atoms of each element present will be definite and constant, and their ratios can be expressed as integers or simple fractions. This property explains the laws of definite proportions and multiple proportions.

* Chemical properties indicate how a substance can be changed into another substance. “Hydrogen reacts with oxygen to produce water” is a statement of a chemical property. Physical properties are independent of other substances and involve no change in the identity of the compound. Melting and boiling points, conductivity, hardness, and color are physical properties.

Atoms combine with one another to form molecules, which are the smallest units of a substance that have the chemical properties of the substance. Dalton assumed that the simplest form of an element was an atom, while the simplest form of a compound was a molecule; but we shall soon see that this assumption is not quite correct because some elements exist as molecules.

Dalton developed a list of symbols to represent the different atoms. Hydrogen was and oxygen was . Lacking any information to the contrary, he assumed water contained one hydrogen atom and one oxygen atom and was therefore represented as . Fortunately, his system was discarded for one in which the symbol of the element was formed from one or two letters of its name, usually the first one or two. Thus, a hydrogen atom is now represented by H and an oxygen atom by O. A water molecule would have been HO, but we now know that a water molecule contains two hydrogen atoms and one oxygen atom, so it is H2O. The story of how the formula of this simple molecule was determined is fascinating and instructive, and it is presented in the next section.



1.4 ATOMS AND MOLECULES

Dalton recognized that mass was an important property of atoms and molecules, so he introduced the concepts of atomic and molecular mass.* Hydrogen is the lightest of the known atoms, so Dalton assigned it a relative mass of one (no units). He assumed that the atom ratio in water was 1:1, so the reaction of hydrogen with oxygen to produce water was thought to be H + O → HO. He also knew that 8 g of oxygen reacted for each 1 g of hydrogen to produce 9 g of water. Therefore, he reasoned that the mass of one oxygen atom was eight times that of one hydrogen atom, which meant that oxygen had a relative mass of 8. Water, which he assumed was HO, had a relative mass of 1 + 8 = 9.

At about the same time that Dalton was formulating his atomic theory, the French chemist, Joseph Gay-Lussac, was measuring the volumes of reacting gases. In 1808, he published his results, now known as the law of combining volumes.

Law of Combining Volumes: Volumes of reacting gases are in simple whole number ratios.

Experiment showed that the volumes of hydrogen and oxygen that react are in a 2:1 ratio.

* The terms atomic and molecular weights are commonly used, but the numbers represent masses not weights.

2 volumes of hydrogen + 1 volume of oxygen → water The law of combining volumes was soon explained in terms of Dalton’s atomic theory, but the explanation rested on the assumption that equal volumes of gases measured at the same temperature and pressure must contain equal numbers of particles! The volume of hydrogen is twice that of oxygen in the reaction, so it was concluded that a water molecule contained twice as many hydrogen atoms as oxygen atoms. The formula of water had to be H2O. The reaction was then thought to be 2H + O → H2O.

The change in the formula of water meant that the relative masses that Dalton had determined for hydrogen and oxygen were wrong. One oxygen atom was eight times more massive than two hydrogen atoms and the atomic mass scale was changed accordingly. In the new scale, H = 1, O = 16, and H2O = 18. The new scale was still consistent with the observation that 1 gram of hydrogen reacted with 8 grams of oxygen to produce 9 grams of water. However, one gram of hydrogen contained twice as many atoms as did 8 grams of oxygen.

The formula of water and the relative masses of hydrogen and oxygen had finally been determined, but the reaction of hydrogen and oxygen still had something to teach us. Consider that the equation 2H + O → H2O predicts that 2 volumes of hydrogen combine with 1 volume of oxygen to produce 1 volume of water vapor, but experiment was soon to show that the reaction produces 2 volumes of water!


2 volumes of hydrogen + 1 volume of oxygen → 2 volumes of water The apparent dilemma was explained in 1811 by Amadeo Avogadro. His hypothesis became known as Avogadro’s law.

Avogadro’s law: Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

Avogadro reasoned that elements, like compounds, can also occur as molecules, and it was the reaction of hydrogen and oxygen molecules, not atoms, that produced water molecules. His suggestion was contrary to Dalton’s assertion that the atom is the simplest form of an element. Thus, an element is a substance that consists of only one type of atom, and a compound is a substance that contains more than one type of atom.

The relative volume of water produced in the reaction of hydrogen and oxygen could be explained if elementary hydrogen and oxygen each existed as diatomic (two atom) molecules, which means that the balanced reaction should be written as follows:

2H2 + O2 → 2H2O The above chemical equation is consistent with the experimental observations of both mass and volume ratios, and it is the way the reaction is written today. Hydrogen and oxygen are not the only elements to exist as diatomic molecules.

Common diatomic elements: H2, N2, O2, F2, Cl2, Br2, and I2.

Some elemental molecules, such as P4 and S8, contain more than two atoms. The atomic mass scale was constructed by measuring relative masses of combining

substances and assuming or determining the formula of the compound they produced. However, the modern system of atomic masses is not based on the hydrogen atom; rather it is based on the most common form of carbon, called carbon-12, which is assigned an atomic mass of exactly 12. The mass of an oxygen atom is 4/3 that of a carbon-12 atom, so oxygen’s atomic mass is (4/3)(12) = 16.



Example 1.2 Classify the following as elements or compounds and as atoms or molecules.

a) S8 S8 contains only one type of atom and is therefore an element. However, it contains eight chemically bound atoms, so it is also a molecule.

b) Ar Ar contains only one type of atom and is therefore an element. In addition, it contains no chemical bonds and is an atom.

c) N2O5 N2O5 contains two types of atoms, so it is a compound. The nitrogen and oxygen atoms are bound together to form a molecule.

Example 1.3 Balance the following chemical equations.

a) N2 + O2 → N2O5 The number of atoms of each element in a balanced equation are made the same on both sides by placing coefficients in front of each species. Subscripts in the formula must not be changed as that would change the identity of the molecules. Note that N2 and O2 are both diatomic molecules, so the number of oxygen and nitrogen atoms will each be even on the left side if integer coefficients are used. We therefore start by placing a 2 in front of the N2O5 to assure an even number of oxygen atoms on the right side.

N2 + O2 → 2N2O5 The right side now shows four nitrogen atoms, which means that two N2 molecules must appear on the left. The right side also indicates ten oxygen atoms, so five O2 molecules are required on the left. Thus, we write

2N2 + 5O2 → 2N2O5 The above equation shows four nitrogen atoms and ten oxygen atoms on each side. The number of each atom is the same on both sides, so the equation is balanced.

b) Al + CuSO4 → Al2(SO4)3 + Cu The S and O atoms remain bound to one another in SO4, so we can balance the SO4 as a unit rather than individual sulfur and oxygen atoms. We start by placing a one in front of the molecule with the greatest number of atoms, Al2(SO4)3. That fixes the number of Al atoms and SO4 units on the right side, so we must balance them on the left as follows:

2Al + 3CuSO4 → 1Al2(SO4)3 + Cu The number of Cu atoms is now fixed on the left at 3, so we balance them on the right. Finally, coefficients of one are not usually shown, so the balanced equation is

2Al + 3CuSO4 → Al2(SO4)3 + 3Cu


1.5 THE MOLE AND MOLAR MASS

The last equation balanced in the preceding exercise shows that Al atoms react with CuSO4 molecules in a 2:3 ratio, so we would like to mix the reactants in a ratio that is close to that. Unfortunately, individual atoms and molecules are much too small to see, and the numbers of atoms and molecules in reactions carried out in the laboratory are far too large to count, so we must measure the ratio of reacting atoms and molecules indirectly from their relative masses and their atomic or molecular masses.

If the masses of two pure substances are in the same ratio as their atomic or molecular masses, then the substances contain the same number of atoms or molecules.

The number of atoms or molecules in 12.0 g C, 16.0 g O, and 18.0 g H2O is identical. While Dalton had no idea what that number was, it was still a very useful concept because chemists could mix ingredients in the desired atom or molecule ratios using masses. The number of atoms or molecules in each of the above examples is called the mole. We count atoms and molecules in moles just as we count our shoes in pairs and our eggs in dozens.

A mole, which is abbreviated mol, is the number of atoms or molecules present in a sample of an element or a compound with a mass equal to its atomic or molecular mass expressed in grams. The number of items in a mole is called Avogadro’s number (NA), which has been determined to be NA = 6.022 x 1023 mol-1.

Just as dozen means 12 items, mole means 6.022x1023 items. Avogadro’s number is a huge number, and its size is an indication of just how small atoms and molecules are. Consider that a mole of dice, each die 1/2 inch on a side, would cover the 48 contiguous states of the United States to a height of 100 miles, while a mole of water has a volume of only 18 mL. Yet, a mole of dice and a mole of water contain the same number of items.

A mass of a pure substance equal to its atomic or molecular mass expressed in grams contains one mole of the substance, so it is referred to as the molar mass (Mm) of the substance. Alternatively, the molar mass equals the atomic or molecular mass expressed in g.mol-1. The atomic mass of O is 16, so its molar mass is 16.0 g.mol-1, while the molecular mass of O2 is 32.0, so its molar mass is 32.0 g.mol-1. The mass of an individual atom or molecule is its atomic or molecular mass expressed in atomic mass units (amu). Thus, the mass of a single oxygen atom is 16.0 amu, the mass of one oxygen molecule is 32.0 amu, and the mass of one water molecule is 18.0 amu. Consequently, there are Avogadro’s number of amu in one gram: 1 g = NA amu = 6.022x1023 amu, which means that 1 amu = (1/NA) g = 1.661x10-24 g.



Example 1.4 a) What is the mass of 0.137 mol CaCO3?

Mass ↔ mole conversions are very important in chemistry, and are most easily done by multiplying the given quantity by the appropriate conversion factor that converts the given quantity into the desired quantity. The molar mass is the conversion factor in this problem.

Mm(CaCO3) = Mm(Ca) + Mm(C) + 3Mm(O) = 40 g/mol + 12 g/mol + 3(16 g/mol)

Mm(CaCO3) = 100 g/mol

We now multiply the given quantity by the conversion factor so that the given units are converted into the desired units (mol CaCO3 cancel). Using units and conversion factors to solve a problem is called the factor label method.

0.137 mol CaCO3 ×100 g CaCO3

1 mol CaCO3 = 13.7 g CaCO3

b) How many moles of CaCO3 are present in 5.36 g of CaCO3?

We determined the molar mass of CaCO3 in Part a, so we multiply the given mass (5.36 g) by the conversion factor (100 g CaCO3/mol CaCO3) such that g CaCO3 cancel and the result is the desired quantity.

5.36 g CaCO3 ×1 mol CaCO3

100 g CaCO3 = 0.0536 mol CaCO3

c) How many moles of oxygen atoms are present in 2.69 g of CaCO3? Two conversion factors must be used in this example: one to convert grams of CaCO3 to moles of CaCO3 and one to convert moles of CaCO3 to moles of oxygen atoms.

2.69 g CaCO3 ×1 mol CaCO3

100 g CaCO3×

3 mol O1 mol CaCO3

= 0.0807 mol O

Note that the units of the denominator of the second conversion factor cancel the units of the numerator of the first conversion factor (mol CaCO3). Using the units of the conversion factors to decide the order and manner of multiplication of the factors in a problem is a powerful tool. Refer to Appendix A for a review of how to use conversion factors in mass ↔ mole conversions.

See Appendix A for more examples.


Ratios of atoms or molecules are equal to ratios of moles of atoms or molecules, and chemists usually use moles rather than numbers of individual particles in the laboratory. The reaction 2H2 + O2 → 2H2O shows us that 2 mol H2 react with every 1 mol O2 and are required to produce every 2 mol H2O. The ratio of the coefficients is the conversion factor, called the stoichiometric factor, that allows us to convert from one substance to another in a balanced chemical equation. The following stoichiometric factors and their reciprocals relate to the reaction of hydrogen and oxygen:

2 2 2

2 2 2

2 mol H 2 mol H 1 mol O; ; 1 mol O 2 mol H O 2 mol H O

Example 1.5 How many moles of H2 are required to react with 8.0 g O2 to produce H2O? First, write the balanced chemical equation for the reaction.

2H2 + O2 → 2H2O

The first thing that must always be done when solving a stoichiometry problem is to covert the masses into moles by dividing by the molar mass, which is 32.0 g.mol-1 for O2.

8.0 g O2 ×

1 mol O232.0 g O2

= 0.25 mol O2

The given number of moles is then converted into the chemically equivalent number of desired moles with the stoichiometric factor that contains both substances. The balanced equation indicates that 2 mol H2 are required for every 1 mol O2, so we write

0.25 mol O2 ×

2 mol H21 mol O2

= 0.50 mol H2 required

This is an example of reaction stoichiometry, the topic of Appendix D.

Chemists seldom add chemicals in the exact stoichiometric ratio, so the amount of product that forms depends upon the amount of the reactant, called the limiting reactant, that is consumed first. Consider the reaction of 5 mol S and 6 mol O2 to produce SO3. There are 5 mol S, and each mole of SO3 requires one mole of S, so there is enough S to produce 5 mol SO3. There are 12 mol O atoms in 6 mol O2, and each mole of SO3 requires 3 mol of O, so there is enough O to make only 4 mol SO3. After 4 mol SO3 are produced, all of the O2 is gone, and no more SO3 can be made. Making 4 mol SO3 requires only 4 mol S, so there is 1 mol S left over. Thus, O2 is the limiting reactant, while S is in excess.



Example 1.6 How much Al2O3 can be produced from 10.0 mol Al and 9.0 mol O2? The balanced chemical equation is 4Al + 3O2 → 2Al2O3.

To find the limiting reactant, we must determine how much Al2O3 can be made from each reactant.

10.0 mol Al × 2 32 mol Al O4 mol Al 2 3

2

= 5.0 mol Al O

9.0 mol O × 2 3

2

2 mol Al O3 mol O 2 3 = 6.0 mol Al O

There is enough Al to make 5.0 mol Al2O3, and enough O2 to make 6.0 mol Al2O3. Therefore, Al is the limiting reactant and 5.0 mol Al2O3 is produced.

How many moles of excess reactant remain? The number of moles of O2 consumed in the reaction with 10.0 mol Al is

× 22

3 mol O10.0 mol Al = 7.5 mol O4 mol Al

7.5 of the 9.0 moles of O2 react, so there are 9.0 – 7.5 = 1.5 mol O2 left over.

Dalton’s atomic theory proved successful in predicting the experimental results of his time and became the accepted way to think about matter. His ideas on atomic masses also proved to be very useful, although some of the atomic masses had to be changed as new data became available. All in all, chemists of the day were quite comfortable with the idea that the smallest unit of matter was the ‘billiard ball’ atom proposed by Dalton. Then, near the end of the 19th century, new and more sophisticated experiments were performed that caused scientists to change their view. However, before we examine these experiments, we need to understand the roles of energy and charge in the study of chemistry.

1.6 ENERGY

Energy plays an important role throughout chemistry. Indeed, there is hardly a chapter in this text in which energy considerations are not required. In simple terms, energy is the capacity to move something. The energy of a substance is the sum of its kinetic energy and its potential energy. Kinetic energy (KE) is energy of motion; an object that is moving has the capacity to make another object move simply by colliding with it. The kinetic energy of a particle of mass m moving with a velocity v is given in Equation 1.1.


KE = 1/2 mv2 Eq. 1.1

If mass is expressed in kg and speed in m⋅s-1, then the kinetic energy is in joules (J). Potential energy is energy due to position. Some examples of objects with potential

energy are a truck at the top of a hill, a stretched rubber band, and a stick of dynamite. In each case, the potential energy can be converted into kinetic energy. For example, releasing the brake of the truck causes the truck to gain speed as it rolls down the hill; and, because it is moving, it has kinetic energy. The stretched rubber band flies (moves) across the room as soon as it is released. The potential energy stored in a stick of dynamite is the result of the relative positions of the atoms in the molecules; that is, the energy is stored in the chemical bonds. The potential energy stored in the chemical bonds of the molecules in the dynamite is transformed into kinetic energy during the explosion.

A change in energy is represented by ΔΕ. The sign of the energy change is a significant consideration, so it is important to calculate it in the same manner each time. The convention used is that energy change equals the final energy, Ef, minus the initial energy, Ei, as shown in Equation 1.2.

ΔE = final energy - initial energy = Ef - Ei Eq. 1.2

ΔE > 0 means that the energy of the object increases, while ΔE < 0 means that the energy of the object decreases.


Example 1.7 a) What is the kinetic energy of a 2200 lb car moving at 40 mph?

This example serves as a basis by which to compare other energies we determine throughout the text. Clearly, the typical car moving at 40 mph has the capacity to move many objects, so it is representative of an object with substantial kinetic energy. All other energy considerations in this text are in joules, so we will calculate this kinetic energy in joules as well. Consequently, the mass must be in kilograms, and the speed must be in meters per second. We use the following equalities 1 kg = 2.2 lb and 1 km = 0.62 mi and the conversion-factor method to obtain the desired units.

2200 lb ×1 kg2.2 lb

= 1000 kg

40 mihr

×1 hr

3600 s×

1 km0.62 mi

×1000 m1 km

= 18 m ⋅s-1

Equation 1.1 can now be applied to obtain the kinetic energy of n object with this mass and speed.



KE = 1

2mv2 = 1

2(1000 kg)(18 m ⋅s-1)2 = 1.6 ×105 J = 160 kJ

How does this energy compare to the energy released in chemical reactions? To obtain 160 kJ of energy, one would have to burn only about 3 g of gasoline or 10 g of sugar.

b) What energy change would the car undergo when it stops? The final kinetic energy is zero, and the initial kinetic energy is 160 kJ.

ΔE = Ef - Ei = 0 - 160 kJ = -160 kJ

Thus, ΔE indicates that the car loses 160 kJ of energy.

Why does the truck roll down hill, the rubber band fly, and the dynamite explode? All three processes occur because systems in nature seek the position of lowest energy; i.e., nature favors processes for which ΔE < 0.* Chemical processes can be understood in terms of this fundamental tendency and the fact that energy changes in chemistry are the result of interactions between charged particles. The relationship between energy and the charge on interacting particles is the topic of the next section.

1.7 ELECTROMAGNETISM AND COULOMB’S LAW

A great deal of research in the 1800’s was centered on electricity and magnetism. Scientists recognized that there was a force of interaction, called the electromagnetic force, between charged particles when they were brought close to one another. The force could be either attractive or repulsive. Charles Augustus Coulomb measured this force and stated his observations in what is now called Coulomb’s law.

Coulomb’s Law: Two particles of charge q1 and q2 separated by a distance r experience a force F as shown in Equation 1.3.

1 22

kq qF = rε

Eq. 1.3

k = 8.9875x109 N·m2/C2 and is called Coulomb’s constant. q1 and q2 are the charges in Coulombs, and ε is the dielectric constant of the medium separating the charges. The dielectric constant is a measure of how well the medium insulates the two charges. ε = 1 in a vacuum, but it is 79 in water, which means that the interaction between charges in water is only about 1.3% of that in a vacuum for the same charges and separation. This decrease in the interaction results because the intervening water molecules have the ability to effectively shield the two charges from one another. When F is negative (q1 and q2 have

* As we will see in Chapter 9, ΔE < 0 is indeed an important driving force, but it is not the only one.


opposite signs), the force is attractive. When F is positive (q1 and q2 have the same signs), the force is repulsive. This is summarized by the statement that opposite charges attract while like charges repel. The force of attraction or repulsion between two particles is very small at large separation but gets stronger as the distance between the particles decreases. The force also becomes stronger as the charges on the particles increase.

In Section 1.5, energy was defined as the capacity to move something, but in order for something to be moved, a force (push or pull) must be exerted. Energy is defined as a force exerted through a distance (E = F x r). Consequently, the change in potential energy that results when two charged particles interact can be obtained by multiplying Equation 1.3 by r. The result is given by Equation 1.4:

1 2

kq qE = rε

Eq. 1.4

E is the potential energy of the two particles separated by distance r relative to their potential energy when they are not interacting* (r = ∞). E, which we will call the energy of interaction between the two particles, depends upon the charges on the particles (q), the distance between them (r), and the intervening medium (ε). A graph of the energy of interaction of two charged particles as a function of the distance between them is shown in Figure 1.1. At large separations, the charged particles do not interact and E ~ 0. As their separation decreases, however, their energy of interaction changes. If the charges are of the same sign, their energy of interaction increases (red line), but if the charges are of opposite sign, their energy of interaction (green line) becomes increasingly negative. Thus, particles of opposite charge are attracted because their potential energy decreases as they get closer to one another, while particles with charges of the same sign are repelled because their potential energy decreases as they move apart.

* E is actually the difference between the energy of interaction at a separation r and the energy when the charges at infinite separation; i.e., ΔE = E(r) - E(∞), but E(∞) = 0, so ΔE = E(r) - 0 = E(r). Hence, the Δ is often dropped and the energy of interaction is simply expressed as E.

r

�E

like charges repelbecause moving apart

lowers their energy

opposite charges attractbecause moving closer

lowers their energy

0

Figure 1.1 Electrostatic or Coulombic energy of interaction Opposite charges are attracted because their energy decreases as they get closer to one another (green line). Like charges repel because moving apart lowers their energy (red line).

Coulomb’s law, in combination with the fact that systems seek the position of lowest energy, is exceedingly important in the study of chemistry because interactions in chemistry can be viewed as interactions between charged particles. We conclude that particles of opposite charge move closer and particles of like charge move apart to minimize their energy. If particles of like charge are forced together or particles of opposite charge are pulled apart, the potential energy of the system rises and the process is unfavorable.



Example 1.8 a) The particles shown in the margin are in the same medium and the numbers are

their relative charges. Arrange the particles from most attractive to most repulsive. The relative forces can be obtained from F ∝ q1q2/r2 because both k and ε in Equation 1.3 are constant in this example. The units of q and r are not important so long as they do not change from one system to the next.

∝ ∝ ∝ ∝2 2 2 2a cb d

(+1)(-2) (+3)(+2) (+3)(-3) (-3)(-2)F = -2; F = +0.67; F = -0.56; F = +1.5

(1) (3) (4) (2)

1+ 2-

1 nm

3+ 3-

4 nm

a)

3+

3-

2+

2-

3 nm

2 nm

b)

d)

c)

Example 1.8

Negative values are attractive, so the order from most attractive to most repulsive is (most negative) a, c, b, d (most positive)

b) Arrange the systems in order of energy of interaction. Use E ∝ q1q2/r for the reasons outlined in Part a to find the relative energies

∝ ∝ ∝ ∝a cb d(+1)(-2) (+3)(+2) (+3)(-3) (-3)(-2)

E = -2; E = +2; E = -2.25; E = +3(1) (3) (4) (2)

(a)

(b)

ZnS-coating onend of tube glows

where cathoderay strikes

CathodeAnode

Evacuated tube

Hole inanode

(+)

(+)

Negatively chargedmetal plate

Positively chargedmetal plate (+)

(-)

(-)

(-)

Spot is moved byas beam is deflected

�

Spot wherebeam strikes

PowerSupply

PowerSupply

�

Figure 1.2 Schematic of a cathode-ray tube (a) In the absence of an electric or magnetic field, the ‘ray’ travels in a straight line from the cathode to the end of the tube where it strikes ZnS, causing it to glow. (b) The ‘ray’ is bent toward the + plate by an electric field. The amount of deflection is given by Δ.

(most attractive) c , a, b, d (most repulsive)

The charges in system a have the lowest potential (interaction) energy.

1.8 ATOMIC STRUCTURE J. J. THOMSON AND THE CATHODE RAY (1897)

By the late 1800’s, electricity had been observed and studied by many scientists, but its origin and nature were not yet understood. One observation made during this period was that radiation was produced by the application of a high voltage across two metallic plates sealed in an evacuated glass tube like the one shown in Figure 1.2. The rays were named cathode rays because they appeared to originate at the cathode or negatively charged plate. Cathode rays are invisible by themselves, but when they strike certain materials, like glass or zinc sulfide (ZnS), they cause the material to glow. When a zinc sulfide coating was applied to the walls opposite the cathode, the surface glowed brightly where the ‘cathode rays’ struck (Figure 1.2a). From the position of the glow, it was clear that the ‘cathode rays’ traveled in straight lines. The apparatus is now called a cathode ray tube or CRT. In 1897, the British physicist J. J. Thomson explored the nature of the ‘rays’ by passing them through external electric and magnetic fields (Figure 1.2b). He made the following observations and conclusions:


Observation: The rays were deflected by the electric and magnetic fields.

Conclusion: Cathode rays were not light rays, because light is not deflected by electric or magnetic fields. He concluded that the rays had to be charged particles.

Observation: The direction of the deflection was away from the negative plate and toward the positive plate.

Conclusion: The particles were negatively charged. Observation: The extent of the deflection (Δ in Figure 1.2) varies directly with the strength

of the applied external field and the magnitude of the negative charge on the particles, but inversely with the mass of the particle (a bowling ball would be deflected far less than a Ping-Pong® ball when struck with the same force).

Conclusion: By measuring how far the particle was deflected, Thomson determined that the charge-to-mass ratio of the particle was q/m = -1.76x1011 C.kg-1.*

The charge-to-mass ratio (q/m) as determined by Thomson was much larger than had been determined for any other charged particle known at the time. In order for the ratio to be so large, either q (the charge) was very large, or m (the mass) was very small. He assumed that the magnitude of the charge on the particles could not be that much greater than that measured for other charged particles, which meant that the mass of the particles must be much smaller. Indeed, he estimated the mass to be less than 1/1000th the mass of the hydrogen atom; contrary to Dalton’s atomic theory, a hydrogen atom was not the smallest particle. Thomson was shocked! These negatively charged particles were later named ‘electrons’ because they were found to be the source of electricity.

* The minus sign results because the charge is negative.

Figure 1.3 Millikan’s oil drop experiment

1. Fine mist of oil is sprayedinto oil drum apparatus

2. Oil dropletsfall throughhole in plate

3. X-rays knockelectrons fromsurrounding air,which stick todroplet

4. Electrically chargedplates influence

droplet's motion

X-ray Source

(+)

( )

5. Observer timesdroplet's motionand controlselectric field

droplet

Although Thomson had determined the charge-to-mass ratio and estimated a probable mass for this new particle, another experiment was needed to determine the charge and mass separately. This experiment would be performed fourteen years later. R. A. MILLIKAN AND THE CHARGE OF THE ELECTRON (1909)

Robert Millikan, an American physicist, was the first to determine accurately the charge on the electron. In an experiment at the University of Chicago, he welded two plates into an oil drum (Figure 1.3). The upper plate had a small hole drilled into it. Above the hole, he created a fine mist of oil droplets. Individual oil droplets passed randomly through the hole one-by-one to descend toward the lower plate. He used a microscope to observe the fall of an individual droplet. Initially, the drop accelerated due to gravity, but eventually the resistance due to the air stopped its acceleration and the drop began to fall at a constant speed called its terminal speed. He determined the droplet's terminal speed and used it to



determine the mass of the droplet. He then fired X-rays into the drum, which removed electrons from some of the molecules in the air. Some of the released electrons attached to the oil droplet creating a negative charge (q) on its surface. Next, he applied an electric field (E) across the two plates creating a positive charge on the top plate and a negative charge on the bottom plate. As the electric field increased, the rate of descent of the droplet slowed as it was increasingly attracted to the positive upper plate and repelled by the lower plate.

Millikan adjusted the voltage across the two plates until the droplet became suspended, moving neither up nor down. At this point, the electrostatic force (qE) that pulled the droplet up equaled the gravitational force (mg*) that pulled it down, so q = mg/E . E, m, and g were all known, so he was able to determine the charge on the droplet. Various experiments yielded different values of q for different droplets, but all of the measured charges were multiples of the same number, -1.6x10-19 C†. Millikan reasoned that the charges on the droplets were different because each droplet had a different number of electrons; i.e., q = nqe, where n is the number of electrons and qe is the charge on each electron. In this way he was able to determine that the charge on an electron must be qe = -1.6x10-19 C. He then used Thomson’s charge-to-mass ratio q/m = -1.76x1011 C.kg-1, and his charge determine that the mass of the electron was me = -1.6x10-19 C/-1.76x1011 C.kg-1 = 9.1x10-31 kg, which is approximately 1/1800 the mass of the hydrogen atom.

Clearly, it was time to refine Dalton’s atomic model. One proposed model was the ‘raisin pudding’ model of Lord Kelvin and J. J. Thomson. They reasoned that because atoms are uncharged themselves, they must contain enough positive charge to balance the negative charge of the electrons. In the Kelvin-Thomson model, the atom resembled raisin pudding with the negatively-charged electrons (the raisins) embedded in a mass of diffuse positive charge (the pudding). E. RUTHERFORD AND THE NUCLEAR ATOM (1911)

A new model called for a new test, and the test of the Kelvin-Thomson model was not long in coming. In another classic experiment, a British physicist, Ernest Rutherford, tested the ‘raisin pudding’ model by bombarding a very thin gold foil with alpha (α) particles (particles with atomic masses of four and charges of +2) moving at 10,000 mi/s (Figure 1.4). If the Kelvin-Thomson model was correct and the positive charge in the atom was diffuse, then these particles should pass straight through the foil with, at most, only minor deflections.

* g is the acceleration due to gravity. g = 9.8 m.s-2 on earth.

† C is the abbreviation for the Coulomb, the basic unit of electrical charge.

Gold Foil

CircularFluorescentScreen

Scattered

particles� Most particlesare undeflected

Source of

particles�

Beam of

particles�

1 in 20,000 particlesdeflect at acute angles

Figure 1.4 Rutherford’s ‘gold foil’ experiment A thin gold foil was bombarded with positively charged α-particles. The path of the particles (red lines) through the foil was determined by noting where the particles struck a fluorescent screen.

Indeed, Rutherford observed that most of the α-particles did pass through the gold foil, and some were deflected slightly, yet one in 20,000 was deflected at a sharp angle. This one particle in 20,000 was totally unexpected, but it provided the information that Rutherford needed. “It was as if you fired a 15-inch shell at a piece of tissue paper, and it came back to hit you!” he would later exclaim. Rutherford proposed a new model of the atom that was consistent with his observations.

1in

20

,00

0p

art

icle

sd

efl

ec

ted

at

as

ha

rpa

ng

le

some particles aredeflected at smaller anglesgold atom

trajectory of

one particle�

mo

st

pa

rtic

les

are

un

de

fle

cte

d

nucleus

Figure 1.5 Rutherford’s model of the atom used to explain the gold foil experiment The positive charge and most of the mass resides in the nucleus, which is represented as a small black dot in the center of the atom. Particles that were deflected at acute angles collided with the nucleus. If drawn to proper scale, the nucleus would be too small to be seen.

Observation: Most of the α-particles passed through undeflected. Conclusion: Most of the volume of the atom is empty space. Observation: Some minor deflections were observed. Conclusion: A very large force would be required to repel these high-energy particles. α-

particles are positively charged, so he assumed that the deflections were due to collisions with another positively-charged and massive particle.

Observation: Only one α-particle in 20,000 experienced acute deflections. Conclusion: Only one α-particle in 20,000 collided with the positively-charged particle in

the atom, so the cross-sectional area of this particle must be less than 1/20,000th that of the atom; yet it must also contain almost all of the mass!

Rutherford’s model of the atom (Figure 1.5) placed the positive charge and essentially all of the mass of the atom in a very small particle, which he called the nucleus. The nucleus occupies almost none of the volume of the atom. He suggested that the electrons orbit the nucleus much as the planets orbit the sun.

Rutherford’s model was another giant step in our understanding of matter, but there were still dramatic changes to come over the next ten years that would revolutionize the way we would think about atoms and molecules. Nevertheless, the Rutherford model had identified an essential feature of the atom: electrons with negative charge surround a nucleus with positive charge. In fact, the electrons in an atom, as well as their interactions with the electrons and nuclei of other atoms, are the focus of chemistry. The electrons, their location about the nucleus, and their number relative to the number of positive charges in the nucleus dictate the chemistry of an element.

To appreciate just how small and dense the nucleus is, consider the following two analogies:

• If the nucleus of an atom were the size of a baseball, the atom would be a sphere with a one-mile diameter.

• A matchbox filled with nuclei (not atoms) would weigh 2.5 billion tons.



1.9 SUBATOMIC PARTICLES, ISOTOPES, AND IONS

As summarized in Table 1.1, scientists have discovered that atoms are composed of three subatomic particles: protons, neutrons, and electrons. The number of protons identifies the nucleus, and the number of protons and neutrons determines its mass.

Atomic number, Z, is the number of protons in the nucleus. It is the number that characterizes an element. If the atomic number of an element is known, then the identity of the element is known and vice versa.

Mass number, A, is the sum of the number of protons and the number of neutrons in the nucleus. Because the mass of each of these particles is nearly 1 amu, the mass number is an integer that closely approximates the nuclear mass in atomic mass units (amu).

The atomic mass is sometimes included with the symbol of the element. For example, the notation 63Cu (read copper-63) indicates that the mass number of Cu is 63. There is no reason to include Z because it is implied in the element’s symbol. Cu means Z = 29 because Cu always has 29 protons. Note that the number of neutrons is A minus Z. Thus, 63Cu has 63 - 29 = 34 neutrons.*

Isotopes are different forms of the same element that have the same atomic numbers (Z) but different masses (A). Consequently, they differ only in the number of neutrons (A - Z). For example: 35Cl and 37Cl, chlorine-35 and chlorine-37, are two isotopes of chlorine. Both contain 17 protons because both are chlorine atoms. They differ in that 35Cl contains 18 neutrons (35 - 17) while 37Cl contains 20 neutrons (37 - 17). Naturally occurring chlorine is 75.8% 35Cl and 24.2% 37Cl. It is this combination of the two that gives a mole of chlorine a mass of 35.5 grams (the molar mass of chlorine).

Table 1.1 Subatomic particles Particle Mass (amu) Charge†

electron 5.49 x 10-4 -1

proton 1.00728 +1

neutron 1.00867 0

† Charge at the atomic and molecular level is usually expressed as multiples of the fundamental unit of charge as determined by Millikan. Thus, the charge on the electron is designated as -1, which means that it carries a charge of (-1)(1.6 x 10-19 C). The charge on the proton is then (+1)(1.6 x 10-19 C ).

* A and Z are sometimes shown in the atom’s symbol. Thus, the element whose symbol is X and whose atomic number and mass are Z and A, respectively would be represented as follows:

ZAX ⇒ # protons

(# protons + # neutrons)X

This notation is redundant because the symbol (X) identifies the element and its atomic number (Z). However, the notation is used when balancing nuclear equations.

Charge is not continuous; it comes in bundles§ of 1.6x10-19 C. Electrons are bundles of negative charge and protons are bundles of positive charge. All charge is comprised of an integral number of these bundles, and the net charge on a particle at the atomic and molecular level is usually given as the difference between the positive and negative charge carriers rather than the actual charge in coulombs.

§ Quantities that are not continuous but come in discrete units are said to be quantized, and the bundles that carry the units are called quanta. Thus, electrons and protons are quanta of negative and positive electrical charge. In Chapter 2, we show that electromagnetic energy is also quantized.

charge on particle = number of protons – number of electrons

Atoms have no charge because the number of protons and electrons are equal. Thus, Al (Z=13) contains 13 protons, 13 electrons, and no net charge. While the number of protons in an atom is fixed, electrons can be added or removed to produce charged particles called ions. Positively charged ions, which are called cations, are produced when electrons are lost, and negatively charged ions, which are called anions, are produced when electrons


are gained. Al3+ is a cation with a charge of +3. An Al atom must lose three electrons to form Al3+, so Al3+ contains 13 protons and 10 electrons. Similarly, an O atom has 8 protons, so the O2- anion must have 10 electrons.†

† Note: +1 and -1 charges are normally indicated as simply ‘+’ and ‘-’ as in Na+ and Cl-. However, the charge is easily overlooked when written in this manner, so 1+ and 1- are used for emphasis in this book. Thus, Na1+ = Na+ and Cll- = Cl-.

Example 1.9

H

He

Li

Be

B

C

N

O

F Ne

Na

Mg

Al

Si

P

S

Cl Ar

K

Ca

Meltin

gpoin

t s(

C)

o

-1000

0

1000

2000

3000

4000

Order number

Figure 1.6 Periodic behavior of a physical property The melting points of the first 20 elements in order of increasing atomic mass. Atoms that are shown in the same color are in the same chemical group.

order number

0

1

2

3

H

He

Li

Be

B

C

N

O

F

Ne

Na

Mg

Al

Si

P

S

Cl

Ar

K

Ca

Ma

xim

um

nu

mb

er

of

Oa

tom

sp

er

ato

m

Figure 1.7 Periodic behavior of a chemical property The maximum number of oxygen atoms that bind to an atom is given as a function of the order number of the atom. Elements shown in the same color, such as N (N2O5) and P (P2O5), are in the same chemical group, while elements between the atoms in red are in the same chemical period.

Indicate whether the energy of the system would increase, decrease, or remain the same during each of the following processes.

a) An anion approaches a cation. An anion is negatively charged and a cation is positively charged. According to Coulomb’s law (Equation 2.4) opposite charges lower their energy when they approach.

b) One cation is moves away from another. Cations are positively charged. Moving like charges away from one another lowers their energy.

c) A neutron moves toward an electron. Neutrons carry no charge, so there is no interaction and the energy remains the same.

1.10 DIMITRI MENDELEEV AND THE PERIODIC LAW (1869)

Well after Dalton’s atomic theory had been accepted, but before Thomson’s experiments, Dimitri Mendeleev, a Russian chemist, was writing a chemistry book describing the reactions and properties of the elements. However, he could not decide on a good way to organize the chapters. As he sought to arrange and label the elements in an orderly fashion, he decided to use their atomic masses because that was the only known sequential number that was available. Consequently, he arranged the elements in order of increasing atomic masses. As Mendeleev studied the elements in this arrangement, he noticed that their chemical and physical properties seemed to cycle (Figures 1.6 and 1.7).

He arranged the elements in rows of a length such that elements of similar properties fell directly beneath one another to form chemical families or groups (see the caption to Figure 1.6). The rows are referred to as periods (see the caption to Figure 1.7). Thus, the properties of the elements change gradually as you progress across a period, but they remain similar as you progress down a group. The elements on the left side of any period tend to be lustrous solids (have a luster or shine) that are malleable (can be beaten into a form), ductile (capable of being drawn into wires), and good conductors of heat and



electricity. These are the typical properties of metals. The elements on the right side of a period tend to be gases or dull, brittle solids that are poor conductors of electricity or heat. These are the typical properties of nonmetals. Thus, the properties of the elements gradually go from metallic on the left side of a period to nonmetallic on the right side of a period. Those elements that have some properties of both metals and nonmetals and lie between the two broader classes are called metalloids.

Mendeleev published his observations as the periodic law in 1869: The elements, if arranged in an order that closely approximates that of their atomic masses, exhibit an obvious periodicity in their properties. The phrase in italics was used because Mendeleev had so much confidence in his periodic law that he reversed the order of some elements in order to maintain the periodicity even though it meant reversing the order of their atomic masses. Thus, tellurium with an atomic mass of 128 was placed before iodine, which has an atomic mass of 127. He did not understand why the properties of the elements did not exactly follow the order of the atomic masses, but he was confident that the concept of periodicity was correct!

Had Mendeleev blindly used a rigid atomic mass ordering of the elements, he also would have placed arsenic (atomic mass = 75) next to zinc (atomic mass = 65) because gallium and germanium had not yet been discovered. This would have placed arsenic directly under aluminum, yet he knew their properties were quite different. Instead, he placed arsenic directly under phosphorus because arsenic and phosphorus have similar properties. Thus, he left two holes in his chart, which he correctly predicted would someday be filled by two new elements. He even went so far as to predict the properties of these two new elements. While other scientists were defining similar ordering schemes for the elements, it was Mendeleev’s faith in the underlying principle of periodicity that made him leave room in his table for undiscovered elements. The confidence displayed by Mendeleev set him above others who were trying to construct similar relationships. For this reason, he is given credit for the periodic law.

We now realize that the parameter that characterizes an atom is its atomic number (Z), but Mendeleev did not know about electrons or protons, so he had no atomic number to reference. He had only the atomic mass. We now understand why tellurium comes before iodine in the periodic order. Although tellurium is heavier than iodine, its atomic number is smaller. Today, the periodic law is stated as:

Arranged in the order of their atomic numbers, the elements exhibit periodicity in their chemical and physical properties.


Mendeleev’s arrangement of the elements has become known as the periodic chart or the periodic table and is the source of a great deal of information about the physical and chemical properties of the elements. The columns define groups, which consist of elements with similar properties. The rows define periods, which contain elements whose properties change gradually. There has been some controversy over the numbering of the groups in the periodic table. The American method, 1A - 8A and 1B - 8B, and the newer method, which numbers the groups as 1 - 18, are both shown on the periodic tables in this book (Figure 1.8 and the back cover). Both methods are in use today, but the older method is still more common, and we will use the older method in the remainder of this text.

The elements in the groups 1A - 8A, the ‘A’ groups, are the main group elements, and groups 1B - 8B are the transition metals. The two rows at the bottom of the periodic table are the lanthanides and the actinides, which are also known as the inner transition metals.

The periodic table contains a great deal of information about the elements. Each box in the table presented in Figure 1.8 contains (from top to bottom) the atomic number, the chemical symbol, and the atomic mass of the element. As shown in the margin, the atomic number of hydrogen is 1, its symbol is H, and its atomic mass is 1.0079.

The properties of an element are dictated by its position in the periodic table. In the table on the following page, they are also indicated by the background color as follows:

The metals occupy the left side of the periodic table. They represent about 75% of the elements. All metals are solids under ordinary conditions, with the exception of mercury (Hg), which is a liquid, and most are lustrous. Metal atoms lose electrons to form cations in their compounds with nonmetals.

The nonmetals lie on the right side of the Periodic Table and represent about 20% of the elements. Many of the nonmetals occur as gases (nitrogen, oxygen, fluorine, chlorine, and the 8A’s); only bromine is a liquid at room conditions, and the remaining five (carbon, phosphorus, sulfur, selenium and iodine) are solids. Nonmetal atoms gain electrons to become anions in their compounds with metals.

The metalloids or semimetals lie on a diagonal between the metals and nonmetals and have characteristics that are intermediate between metals and nonmetals. Metalloids are shiny like metals, but brittle rather than malleable. Metalloids are neither good conductors nor nonconductors of electricity. Instead they are semiconductors (their conductivity changes with temperature), with silicon being a well-known example.

metal

nonmetal

metalloid

1 H

1.0079

Atomic Number

Atomic Symbol

Atomic Mass (Weight)



56

Ba

137.33

56

Ba

137.33

1

H

1.0079

1

H

1.0079

2

He

4.0026

2

He

4.0026

5

B

10.811

5

B

10.811

6

C

12.011

6

C

12.011

7

N

14.007

7

N

14.007

8

O

15.999

8

O

15.999

9

F

18.998

9

F

18.998

10

Ne

20.180

10

Ne

20.18013

Al

26.982

13

Al

26.982

14

Si

28.086

14

Si

28.086

15

P

30.974

15

P

30.974

16

S

32.066

16

S

32.066

17

Cl

35.453

17

Cl

35.453

18

Ar

39.948

18

Ar

39.948

3

Li

6.941

3

Li

6.941

4

Be

9.0122

4

Be

9.012211

Na

22.990

11

Na

22.990

12

Mg

24.305

12

Mg

24.305

19

K

39.098

19

K

39.098

20

Ca

40.078

20

Ca

40.078

21

Sc

44.956

21

Sc

44.956

22

Ti

47.88

22

Ti

47.88

55

Cs

132.91

55

Cs

132.91

39

Y

88.906

39

Y

88.906

40

Zr

91.224

40

Zr

91.224

23

V

50.942

23

V

50.942

41

Nb

92.906

41

Nb

92.906

24

Cr

51.996

24

Cr

51.996

42

Mo

95.94

42

Mo

95.94

87

Fr

(223)

87

Fr

(223)

88

Ra

226.03

88

Ra

226.03

37

Rb

85.478

37

Rb

85.478

38

Sr

87.62

38

Sr

87.62

57

La

138.91

57

La

138.91

72

Hf

178.49

72

Hf

178.49

73

Ta

180.95

73

Ta

180.95

74

W

183.84

74

W

183.84

89

Ac

227.03

89

Ac

227.03

104

Rf

(261)

104

Rf

(261)

105

Db

(262)

105

Db

(262)

106

Sg

(263)

106

Sg

(263)

25

Mn

54.938

25

Mn

54.938

26

Fe

55.847

26

Fe

55.847

27

Co

58.933

27

Co

58.933

29

Cu

63.546

29

Cu

63.546

30

Zn

65.39

30

Zn

65.39

31

Ga

69.723

31

Ga

69.723

32

Ge

72.61

32

Ge

72.61

33

As

74.922

33

As

74.922

34

Se

78.96

34

Se

78.96

35

Br

79.904

35

Br

79.904

36

Kr

83.80

36

Kr

83.80

49

In

114.82

49

In

114.82

81

Tl

204.38

81

Tl

204.38

50

Sn

118.71

50

Sn

118.71

51

Sb

121.75

51

Sb

121.75

52

Te

127.60

52

Te

127.60

53

I

126.90

53

I

126.90

54

Xe

131.29

54

Xe

131.29

82

Pb

207.2

82

Pb

207.2

83

Bi

208.98

83

Bi

208.98

84

Po

(209)

84

Po

(209)

85

At

(210)

85

At

(210)

86

Rn

(222)

86

Rn

(222)

43

Tc

(98)

43

Tc

(98)

44

Ru

101.07

44

Ru

101.07

45

Rh

102.91

45

Rh

102.91

46

Pd

106.42

46

Pd

106.42

47

Ag

107.87

47

Ag

107.87

28

Ni

58.693

28

Ni

58.693

48

Cd

112.41

48

Cd

112.41

75

Re

186.21

75

Re

186.21

76

Os

190.23

76

Os

190.23

77

Ir

192.22

77

Ir

192.22

78

Pt

195.08

78

Pt

195.08

79

Au

196.97

79

Au

196.97

80

Hg

200.59

80

Hg

200.59

107

Bh

(264)

107

Bh

(264)

108

Hs

(265)

108

Hs

(265)

109

Mt

(268)

109

Mt

(268)

110

Uun

(269)

110

Uun

(269)

111

Uuu

(272)

111

Uuu

(272)

112

Uub

(277)

112

Uub

(277)

58

Ce

140.11

58

Ce

140.11

59

Pr

140.91

59

Pr

140.91

60

Nd

144.24

60

Nd

144.24

61

Pm

(145)

61

Pm

(145)

62

Sm

150.36

62

Sm

150.36

63

Eu

151.97

63

Eu

151.97

64

Gd

157.25

64

Gd

157.25

66

Dy

162.5

66

Dy

162.5

67

Ho

164.93

67

Ho

164.93

65

Tb

158.93

65

Tb

158.93

68

Er

167.26

68

Er

167.26

69

Tm

168.93

69

Tm

168.93

70

Yb

173.04

70

Yb

173.04

71

Lu

174.97

71

Lu

174.97

90

Th

232.04

90

Th

232.04

91

Pa

231.04

91

Pa

231.04

92

U

238.03

92

U

238.03

93

Np

237.05

93

Np

237.05

94

Pu

(244)

94

Pu

(244)

95

Am

(243)

95

Am

(243)

96

Cm

(247)

96

Cm

(247)

97

Bk

(247)

97

Bk

(247)

98

Cf

(251)

98

Cf

(251)

99

Es

(252)

99

Es

(252)

100

Fm

(257)

100

Fm

(257)

101

Md

(258)

101

Md

(258)

102

No

(259)

102

No

(259)

103

Lr

(262)

103

Lr

(262)

Figure 1.8 Modern periodic table See text on previous page for descriptions of metals, metalloids, and nonmetals


The chemical properties of an element can be determined from its group:

Hydrogen is unique as shown by its position in the periodic table in Figure 1.8. It normally forms a +1 ion, which would put it into Group 1A, where it is placed in the periodic table on the back cover. However, it is a nonmetal and can also form compounds in which it is a -1 ion, which is more consistent with the Group 7A elements.

Group 1A elements are known as the alkali metals. All members of this family are very reactive, readily losing one electron to form +1 ions.

Group 2A elements are known as the alkaline earth metals. They are also very reactive but not as reactive as the 1A’s. They occur as +2 ions in their compounds.

Groups 1B - 8B elements are the transition metals. Most form a +2 ion, but +1 and +3 ions are also found. Unlike the 1A and 2A metals, many transition metals can form more than one type of ion, such as Fe2+ and Fe3+.

Group 3A elements are metals, except for boron, which is a metalloid. They typically form +3 ions. Thallium (Tl) forms both +3 and +1 ions.

Group 4A elements demonstrate that metallic character also increases in going down a group. Carbon is a nonmetal and does not typically form ions. Silicon and germanium are metalloids, but tin and lead are metals that form +2 ions.

Group 5A elements also progress from nonmetallic at the top to metallic at the bottom. Nitrogen and phosphorus are nonmetals, which can form -3 ions. Arsenic, antimony, and bismuth increase in their tendency to form +3 ions.

Group 6A elements display only gradual changes in their chemistry, except for the chemistry of oxygen, which differs substantially from the other members. Oxygen, sulfur, and selenium are nonmetals; tellurium is a metalloid, and polonium is a radioactive metal.

Group 7A elements are known as the halogens. They are very reactive, readily forming -1 ions. In fact, they are found in nature only in compounds such as NaCl and MgF2.

Group 8A elements are known as the noble gases because they show little tendency to react with other substances. There are no known compounds of helium, neon or argon and only a very few of krypton and xenon. Radon is radioactive.



Example 1.10 Identify each of the atoms or ions below based on the number of protons and electrons they contain.

Protons Electrons

a) 11 10

b) 16 18

c) 26 26

a) This element contains 11 protons, so it must be Na (sodium). The fact that it contains only 10 electrons means that it carries a charge of 11 - 10 = +1. Thus, the species is Na1+, a cation or positively-charged ion.

b) An atomic number of 16 signifies sulfur. It has gained two electrons to carry a charge of 16 - 18 = -2. It is the S2- anion or negatively-charged ion.

c) Fe (iron) has an atomic number of 26 and, because the number of electrons equals the number of protons, this species is the Fe atom.

Example 1.11 How many protons, neutrons, and electrons are in each of the following?

a) 37Cl1- Chlorine’s atomic number is 17, so it has 17 protons. The number of neutrons (N) in an element is the difference between mass number (A) and the atomic number (Z): N = A - Z = 37 - 17 = 20 neutrons. 37Cl1- has a charge of -1. Charge equals (the number of protons) – (the number of electrons), so -1 = 17 – x for this ion. Solving for x, we find that there are x = 18 electrons in the ion.

b) 40Ca2+ Z = 20 for calcium, so it must have 20 protons. A= 40, so N = 40 - 20 = 20 neutrons for this isotope. The +2 charge means that the number of electrons is two less than the number of protons. Thus, there are 18 electrons in this ion.

c) 127I There is no charge, so the number of electrons equals the number of protons. Z = 53, so the atom contains 53 protons and 53 electrons. In addition, A = 127, so this isotope of iodine contains 127 - 53 = 74 neutrons.



1.11 CHAPTER SUMMARY AND OBJECTIVES

Early scientists used the scientific method to study the chemical and physical properties of matter. Their experiments led to the classification of pure substances as elements or compounds. Initially, they identified atoms as very small spheres characterized by mass. By the early 20th century, scientists had arrived at the nuclear model of the atom. In this model, the atom has a very small, dense, positively charged nucleus that contains the protons and neutrons and is orbited by even smaller, negatively charged electrons. The number of protons in the nucleus, which is the atomic number, characterizes the atom, while the number of electrons that surround the nucleus dictates the charge on the species. If the number of electrons exceeds the number of protons, the substance is negatively charged and called an anion. If the number of protons exceeds the number of electrons, the species is positively charged and called a cation.

The periodic table is one of the greatest classifying systems in science. It orders logically a great deal of information about an element’s chemical and physical properties, such as its atomic number, whether it is a metal, nonmetal, or metalloid, and the ease with which it undergoes chemical reaction.

After studying the material presented in this chapter, you should be able to: 1. explain the scientific method (Section 1.1);

2. distinguish between atoms, molecules, elements, and compounds (Sections 1.2 - 1.4);

3. determine molar mass of elements and compounds and determine the number of moles present in a sample (Section 1.5);

4. define kinetic energy and potential energy (Section 1.6);

5. use Coulomb’s law to explain how the energy of two particles varies with the charge of, the distance between, and the medium that separates them (Section 1.7);

6. describe the experiments that led to the discovery of the electron and to the determination of its charge and mass (Section 1.8);

7. describe the nuclear atom and the experiments that led to its discovery (Section 1.8);

8. list the common subatomic particles with their mass numbers and charge; distinguish between atomic number (Z) and mass number (A); and determine the number of protons, neutrons, and electrons in an ion or atom (Section 1.9); and

9. describe the meaning of all numbers and symbols given on the periodic table and determine whether an element is a metal, nonmetal, or metalloid from its position on the periodic table (Section 1.10).



1.12 EXERCISES 1. Distinguish between a qualitative observation and a quantitative

observation. Give two examples of each.

2. Distinguish among the terms element, compound, atom and molecule. Give an example of an element that is a molecule and an example of an element that is an atom.

3. Indicate whether each of the following represents a mixture of atoms, a mixture of molecules, or a mixture of atoms and molecules. Is each a mixture of elements, compounds, or elements and compounds?

(a) (b) (c) (d)

4. Indicate whether each of the following represents a mixture of atoms, a mixture of molecules, or a mixture of atoms and molecules. Is each a mixture of elements, compounds, or elements and compounds?

(a) (b) (c) (d)

5. Classify each of the following as an element or compound: a) P4 b) Fe c) C3H6O d) SO2 e) O3

6. Classify each of the following as an element or compound: a) H2O b) C60 c) Au d) CO e) N2

7. Classify each of the following as an atom or a molecule: a) P4 b) Fe c) C3H6O d) SO2 e) O3

8. Classify each of the following as an atom or a molecule: a) H2O b) C60 c) Au d) CO e) N2

9. Classify each of the following as an atom, molecule, anion, or cation: a) NH3 b) NH4

1+ c) N3- d) CH3COO1- e) Si

10. Classify each of the following as an atom, molecule, anion, or cation: a) Na1+ b) NO3

1- c) Na d) Cl1- e) Al3+

11. Give the name of each of the following elements: a) Na b) Br c) Hg d) Fe e) Ag

12. Give the name of each of the following elements: a) Pb b) Au c) F d) Ca e) P

13. Write the symbol for each of the following elements: a) copper b) lead c) strontium d) silicon e) tin

14. Write the symbol for each of the following elements: a) potassium b) iron c) nickel d) cadmium e) selenium

15. Determine the number of moles of atoms that are present in each of the following samples:

a) 5.0 g K b) 17 g Mg c) 3.0 g C d) 2.2 kg Fe e) 14 mg Ag

16. How many moles are in a 5.0-g sample of each of the following elements? How many moles of atoms are in each sample?

a) Ca b) F2 c) O2 d) S8 e) P4.

17. How many moles of molecules are in a 10.0-g sample of each of the following compounds? How many moles of atoms are in each sample?

a) SF6 b) CCl4 c) C6H14 d) SO3 e) BF3

18. Determine the number of moles of carbon that are present in each of the following samples: a) 1.0 g of aspirin, C9H8O4 b) 3.0 g of ibuprofen, C13H18O2 c) 12 mg of acetaminophen (Tylenol), C8H9NO2

19. Determine the mass of the following samples: a) 2.5 mol CaCl2 b) 0.75 mol C6H12 c) 1.8 mol CO2

20. Determine the mass of the following samples: a) 4.6x10-3 mol Al(NO3)3 b) 3.6 mol C12H22O11 c) 220 mol H2

21. Consider a 5.00 g sample of Ca2S3. a) How many moles of Ca2S3 does it contain? b) How many moles of sulfur does it contain? c) How many grams of sulfur does it contain?

22. A sample of Al(NO3)3 contains 0.90 moles of nitrogen. a) How many moles of Al(NO3)3 are present in the sample? b) What is the mass of the sample in grams? c) How many moles of oxygen are in the sample? d) How many aluminum atoms are in the sample? e) How many grams of aluminum are in the sample?


23. Balance the following equations using the smallest integer coefficients: a) N2 + H2 → NH3 d) CO2 + H2O → C6H12O6 + O2 b) C2H6 + O2 → CO2 + H2O e) P4O10 + H2O → H3PO4 c) Al + O2 → Al2O3 f) Ca + O2 → CaO

24. Balance the following equations using the smallest integer coefficients: a) P4 + H2 → PH3 b) H3PO4 + KOH → K3PO4 + H2O b) Cl2 + O2 → Cl2O7 e) N2O5 + H2O → HNO3 c) Al + S8 → Al2S3 f) Al + H2O → Al(OH)3 + H2

25. Consider the reaction of 0.30 mol K with O2, 4K + O2→ 2K2O a) How many moles of molecular oxygen are required? b) How many moles of potassium oxide would form? c) What mass, in grams, of potassium oxide would form?

26. Consider the reaction of 6.5 g of iron with O2, 4Fe + 3O2 → 2Fe2O3 a) How many moles of iron react? b) How many moles of oxygen must react? c) How many moles of Fe2O3 are formed? d) What is the mass, in grams, of the Fe2O3?

27. The green molecules (G2) react with the yellow molecules (B2) to form G3B as shown to the right. Each circle represents one mole of the atoms. Atomic masses: G = 10 and B = 30. a) How many grams of G2 and B2 are in the container? b) How many G3B molecules can be produced? c) What is the balanced chemical equation for the reaction? d) How many grams of G3B would be produced? e) What mass of G2 or B2 molecules would be left over?

28. The yellow molecules (Y3) react with the blue molecules (B2) to form Y2B as shown to the right. Each circle represents one mole of the atoms. Atomic masses: Y =20 and B = 25. a) How many grams of Y3 and B2 are in the container? b) How many Y2B molecules can be produced? c) What is the balanced chemical equation for the reaction? d) How many grams of Y2B would be produced? e) What mass of Y3 or B2 molecules would be left over?

29. Consider the reaction of 6 mol Fe and 6 mol O2 to produce Fe3O4. a) Write the balanced chemical equation.

b) How many moles of Fe3O4 could be produced? c) How many moles of excess reactant remain after the reaction is done?

30. Consider the reaction of 3 mol P4 and 10 mol O2 to produce P2O5. a) Write the balanced chemical equation. b) How many moles of P2O5 could be produced? c) How many moles of excess reactant remain after the reaction is done?

31. Use Coulomb’s law to explain why Na1+ ions and Cl1- ions exist as separated ions in liquid water (ε = 79) but form ion pairs (NaCl units) in liquid carbon tetrachloride (ε = 2).

32. Explain what happens to the energy of an electron and a proton as the distance between them decreases. Explain how the energy of two protons changes as the distance between them decreases.

33. What is the charge in coulombs of a mole of electrons?

34. List the following systems of charged particles from most negative to most positive energies of interaction. Also list the forces from most attractive to most repulsive. a) +2 and -3 charges separated by 10 nm b) -2 and -1 charges separated by 8 nm c) -2 and +2 charges separated by 8 nm

35. List the following systems of charged particles from most negative to most positive energies of interaction. Also list the forces from most attractive to most repulsive. a) +2 and +2 charges separated by 10 nm b) -2 and +3 charges separated by 11 nm c) +2 and +1 charges separated by 6 nm

36. Consider Thomson’s experiments with cathode rays. a) What conclusion was drawn because the “rays” were deflected by

electric and magnetic fields? b) What information about the rays was deduced from the fact that the

“rays” moved toward the positive plate of the electric field? c) What two factors dictated the extent of deflection of the rays?

37. Why didn’t the oil droplets in Millikan’s experiment all have the same charge? What did the charges all have in common?



47. Write the symbol for the species with the number of protons and electrons shown below.

48. Write the symbol for the species with the number of protons and electrons shown below.

d) the transition metal in the same family as iron and the same period as antimony

49. Distinguish between a group and a period. How are the properties of the elements in each related?

a) 75As5+ b) 31P3- c) 195Pt d) 235U e) 207Pb2+

a) 16O2- b) 27Al3+ c) 25Mg d) 19F e) 48Ti4+

52. Group the following elements in pairs that are likely to have similar chemical properties: Li, N, F, P, K, and Br.

53. Group the following elements in pairs that are likely to have similar chemical properties: Ca, S, Sr, He, O, Ar.

45. Determine the number of protons, neutrons and electrons in

46. Determine the number of protons, neutrons and electrons in

b) the nonmetal in the same family as germanium

a) the alkali metal in the same period as bromine

a) the metalloid in the same family as gallium

c) the noble gas in the same period as silicon

e) the only metal in the same group as sulfur d) the halogen in the same period as lead

a) 30 protons and 28 electrons

a) 34 protons and 36 electrons

51. Identify each of the following elements:

50. Identify each of the following elements:

b) the lightest alkaline earth metal

c) the only gas in the fifth period

c) 47 protons and 47 electrons b) 26 protons and 23 electrons

b) 81 protons and 78 electrons c) 7 protons and 10 electrons

40. The paths of two particles, A and B, which pass between the plates of an electric field designated by ‘+’ and ‘-’, are shown below. One is a cation with a +1 charge, and the other is an anion with a -1 charge. The line in the center represents the path of an uncharged particle.

41. Use Figure 1.7 and the periodic law to determine the formulas of the oxides of the following elements:

c) The fraction of α-particles that was deflected at sharp angles was very small.

a) Rb or Ca b) Si or Sn c) C or Pb d) Cs or Xe e) He or Xe

38. What conclusions did Rutherford draw from the following observations? a) Most α-particles passed through the foil with little or no deflection.

42. Use Figure 1.6 and the periodic nature of the elements to predict which element in each pair has the higher boiling point.

a) phosphorus b) arsenic c) selenium

39. The deflection of an X1+ ion in an electric field is 14% of that of an α-particle. What is the identity of X? An α-particle is 4He2+.

43. Use periodic behavior and the given chemical formulas to predict the formulas of the compound formed between the following elements:

44. Use periodic behavior and the given chemical formulas to predict the formulas of the compound formed between the following elements:

b) What is the approximate ratio of the masses of the two particles? Express your answer as the ratio (mass of B)/(mass of A).

b) Some α-particles were deflected at very sharp angles.

b) Sc and Br, given the formulas KBr and CaBr2 a) Pb and Cl, given the formulas TlCl and BiCl3

a) Identify each particle as an anion or a cation.

b) Na and N, given the formulas NaF and Na2O

-

+ A

B

Source 0

+1

-1

-2

a) Al and S, given the formulas Na2S and MgS

d) carbon e) cesium


Chapter 1 1.0 Introduction 1.6 Energy 1.1 Scientific Method … · 1.4 Atoms and Molecules 1.10...

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Transcript of Chapter 1 1.0 Introduction 1.6 Energy 1.1 Scientific Method … · 1.4 Atoms and Molecules 1.10...